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I need to evaluate $\large\sum_{i=-\infty}^{n} 2^i$.

This is what I tried:

$$S = 2^{-\infty} + \color{red}{2^{-\infty + 1} + 2^{-\infty + 2} + ... + 2^{n-1} + 2^{n}}$$ $$2S = \color{red}{2^{-\infty + 1} + 2^{-\infty + 2} + 2^{-\infty + 3} + ... + 2^{n}} + 2^{n+1}$$

$$ \begin{align*} 2S - S & = (A + 2^{n+1}) - (2^{-\infty} + A)\\ & = 2^{n+1} - 2^{-\infty} \end{align*}$$

Is my process (and answer) correct?

David G
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    No not really... how do you define $-\infty + 1$? Your answer is almost correct: the correct answer is $2^{n+1}$ – Mathmo123 Sep 10 '14 at 19:18

4 Answers4

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Let's re-index things to put them in terms of more familiar series.

By letting $j = -i$, this expression is the same as

$$ \sum_{j = -n}^\infty 2^{-j} $$

If we then let $k = j + n$, then

$$ \sum_{k = 0}^\infty 2^{-k + n} = 2^{n} \sum_{k=0}^\infty 2^{-k} $$

so we're just left computing $\sum_{k=0}^\infty 2^{-k}$ (which converges by e.g. the ratio test). This is a common geometric series, but if you haven't encountered it before, you can determine its value using the same approach as in your original question:

$$ S = 1 + \frac12 + \frac14 + \dots \\ 2S = 2 + 1 + \frac12 + \dots \\ S = 2S - S = 2 + 1 - 1 + \frac12 - \frac12 + \dots = 2 $$

so

$$ 2^{n} \sum_{k=0}^\infty 2^{-k} = 2^n\cdot 2 = 2^{n+1}. $$

BaronVT
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    I get how you simplified to $2^n \sum_{k = 0}^\infty 2^{-k}$, but not how you got $2^{n+1}$ from that... – David G Sep 10 '14 at 19:32
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    Sorry, I've assumed that you know how to sum a geometric series. I'll add some more explanation. – BaronVT Sep 10 '14 at 19:33
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Here is an intuitive explanation; to turn it into a rigourous proof, see below.

Let $$S = 2^n + 2^{n-1} + \cdots$$Then $$\begin{align}2S &= 2^{n+1}+2^n+2^{n-1}+\cdots\\ &= 2^{n+1} + S\end{align}$$

So $S = 2S-S = 2^{n+1}$

Note that in doing this, we don't have to think about what "$2^{-\infty}$" would be; this value isn't strictly defined. This is an infinite series, and as such it has no final term. Given any term, I can create the next term from it by dividing by 2, so if there were a final term, then I could divide by 2 to get another term!

In order to prove this rigourously we must make sure that the series converges, so that $S$ actually has a value that we can manipulate arithmetically. Define $$S_k=\sum_{i=-k}^n2^i$$

Then $$\begin{align} 2S_k&= \sum_{i=-k}^n2^{i+1}\\ &=\sum_{i=-k}^n2^i + 2^{n+1}-2^{-k}\\ &=S_k + 2^{n+1}-2^{-k} \end{align}$$

So $$S_k = 2^{n+1} -2^{-k}$$ and as $k \to \infty$, $2^{-k} \to 0$, so $$S_k \to 2^{n+1}$$

Mathmo123
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Notice that $$\sum_{i=-\infty}^n2^i=\sum_{i=-n}^{\infty}2^{-i}=\sum_{i=-n}^02^{-i}+\sum_{i=1}^{\infty}2^{-i}=\sum_{i=0}^n2^i+\sum_{i=1}^\infty\frac{1}{2^i}$$ This reduces the problem to show that $\sum_{i=1}^\infty\frac{1}{2^i}=1$ which is a very well documented result and that $\sum_{i=0}^n2^i=2^{n+1}-1$ which can be shown with induction.

BeaumontTaz
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Such "infinite sums" are not really sums. They are limits of finite sums. You don't compute them by adding up infinitely many terms--it isn't possible to perform an infinite number of steps in a computation. The symbol

$$S = \sum_{i=-\infty}^n 2^i$$

means

$$S = \lim_{N\rightarrow\infty}\sum_{i=-N}^n 2^i$$

This is computed by first finding the value of

$$S_N \equiv \sum_{i=-N}^n 2^i = 2^{-N} +\cdots + 2^n$$

and then computing the limit

$$S = \lim_{N\rightarrow\infty}S_N$$

To find what $S_N$ is, just note that $$S_N =2S_N -S_N = 2(2^{-N} +\cdots + 2^n)-(2^{-N} +\cdots + 2^n)$$ $$=(2^{-N+1} +\cdots + 2^{n+1}) - (2^{-N} +\cdots + 2^n)$$ $$=2^{n+1} - 2^{-N}$$

Then you can see at once that

$$S = \lim_{N\rightarrow\infty}S_N = \lim_{N\rightarrow\infty}(2^{n+1} - 2^{-N}) = 2^{n+1}$$

MPW
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